Problem: A particle moving in the $xy$ -plane has velocity vector given by $v(t)=\left(-2e^t,e^t\right)$ for time $t\geq 0$. What is the magnitude of the displacement of the particle between time $t=2$ and $t=3$ ? Round to the nearest tenth.
To find the magnitude of the displacement of the particle, we should first find the particle's horizontal displacement $\Delta x$ and the particle's vertical displacement $\Delta y$. Then we can find the magnitude of the displacement using the distance formula: $\text{Magnitude of displacement }=\sqrt{(\Delta x)^2+(\Delta y)^2}$ The particle's horizontal displacement can be found by taking the definite integral of the horizontal component of $v(t)$ between time $t=2$ and $t=3$ : $\Delta x=\int_{2}^{3} -2e^t\,dt=-2e^3+2e^2$ The particle's vertical displacement can be found by taking the definite integral of the vertical component of $v(t)$ between time $t=2$ and $t=3$ : $\Delta y=\int_{2}^{3} e^t\,dt=e^3-e^2$ Now we can find the magnitude of the displacement: $\begin{aligned} &\phantom{=}\sqrt{(\Delta x)^2+(\Delta y)^2} \\\\ &=\sqrt{\left(-2e^3+2e^2\right)^2+\left(e^3-e^2\right)^2} \\\\ &\approx 28.4 \end{aligned}$ In conclusion, the magnitude of the displacement of the particle between time $t=2$ and $t=3$ is $28.4$ units.